Just for fun, I set up the following scheme:
- A 6j symbol is everything that fulfils Biedenharn-Elliott. (Plus symmetry, orthogonality etc. if that doesn't follow from it anyway.)
- There are only a finite number $i$ of irreps (including $1$, the 1-dim "neutral").
Then I tried out which of the many possible Clebsch-Gordan expansions are mutally consistent with all the 6j equations via Biedenharn-Elliott. For two irreps I got only this: $1\bigotimes{X}=X (X=1,2), 2\bigotimes{2}=1+2.$
- Ha!
- Even I did see this before, it's the smallest Fibonacci fusion category. Now a quick research gave me a load of adjectives that come with fusion categories. Can you tell me which of them apply to my scheme?
- $i=3$ produces 3 "minis", with $i=4$ I get 6 (and I stopped here because of the dreaded combinatiorial explosion - are these already classified somewhere?).
- My main question, though, is whether you get a free parameter when you choose $i$ large enough. All my "minis" give only a set of fixed complex numbers for the values of the 6j symbols, the quantum dimension and the writhe normalizer (the minis all seem to be knot-theory compatible). The latter is a root of unity so I can speculate all the minis correspond to special values of the Jones polynomial. Or suchlike.

The number of non-isomorphic simple objects is called the rank of the fusion category.

You've made an error somewhere, as in rank 2 you should also get an example where the nontrivial simple object squares to $1$. (In your notation $2 \otimes 2 \cong 1$.)

For rank 2, Ostrik gave a complete classification (http://arxiv.org/abs/math/0203255). Ostrik's argument is somewhat indirect (via the Drinfel'd center). As far as I know, no one has given a direct classification in rank 2 via 6j symbols. The hard part is figuring out why there's no fusion categories with the fusion rule $X^2 = 1 + n X$ for $n>1$.

A great source of all information about "minis" that give knot invariants is Section 5.3 of Rowell-Strong-Wang (http://arxiv.org/abs/0712.1377). "Modular" means that you get knot invariants plus you have a certain "non-degeneracy" condition that's a bit hard to explain.

Ah, silly me, of course I got 2*2=1 too but the object looked so trivial that I didn't mention it. :-) (Also, I probably have a different counting: the solutions with phi or 1/phi I see as the same.) THX for the literature list. BTW, after I did the computations I noticed that I complete overlooked that possibility: why not X*X=1+nX? Do I infer correctly that multiplicy>1 CAN occur when rank>2? (The K(0,1,0,2) mentioned in arxiv.org/abs/math/0503564, right? I have to check it out with Clebsches, knot incompatible or not :-) @Bruce: That saves me lot of time, THX too!
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Hauke ReddmannAug 22 '12 at 10:08

Indeed, you can have nontrivial multiplicity for rank 3 as you say.
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Noah SnyderAug 22 '12 at 14:27